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Let \(X\) be a discrete r.v. and \(Y = g(X)\). If \(g\) is strictly monotonic, we have \(h(y) = g^{-1}(y) = x\). Thus
\[p_Y(y) = P(y = g(x)) = P(x = h(y)) = p_X(h(y)).\]Let \(X\sim N(\mu, \sigma^2)\) and \(Y = aX+b\). We have
\[Y \sim N(aX+b, a^2\sigma^2).\]Let \(X\) be a continuos r.v. and \(Y = g(X)\). If \(g\) is strictly monotonic, we have \(h(y) = g^{-1}(y) = x\). Moreover,
\[f_Y(y) = f_X(h(y))\Bigg\vert {dh(y) \over y} \Bigg\vert.\]這個等式由 CDF 微分而來。（分成 mono. increasing 和 mono. decreasing 討論，因此有絕對值）
Let \(X, Y\) be independent continous r.v.s, and \(Z = g(X,Y)\). We have
\[F_Z(z) = P(g(X,Y) \le z).\]列出這個式子後，在平面上畫出 \(X\) 和 \(Y\) 有定義的區域，藉由算面積的方式來求 \(F_Z(z)\)！
詳見 MIT OCW。
Generate sample \(X\) with given CDF \(F_X(\cdot)\).
已經知道 CDF 了，利用 uniform distribution \(u\) 求出 \(g(u)\) such that \(x = g(u)\).
為甚麼要這樣做？為了用 uniform distribution 表示其他 distribution？
Let \(X, Y\) be independent, known PMFs/PDFs, and \(Z=X+Y\). Then we have
\[p_Z(z) = \sum_xp_X(x)p_Y(z-x), \\ f_Z(z) = \int^{\infty}_{-\infty}f_X(x)f_Y(z-x)dx.\]這是 convolution！
如果 \(X\) 和 \(Y\) 都是 normal r.v.，則 \(Z\) 也是 normal（代入即可證明），而且 \(Z\sim N(\mu_x+\mu_y, \sigma_x^2 + \sigma_y^2)\)。
Poisson 加起來也還是 Poisson，而且 \(Z\sim \text{Poi}(\lambda_x + \lambda_y)\).
見 tranforms。